x³ + y³ = 22 and x + y = 5 then find the approximate value of x⁴ + y⁴.

Question

Question

Answer 1

We know that

x³ + y³ = (x + y)(x² + y² – xy)

Now we have x³ + y³ = 22 and x + y = 5

⇒ 22 = 5(x² + y² – xy)

⇒ 22 = 5[(x + y)² − 3xy)]

⇒ 22 = 5[(5)² − 3xy)]

⇒ xy = 103/15

Now multiply x³ + y³ = 22 with x + y = 5

⇒ x⁴ + y⁴ + xy(x² + y²) = 110

⇒ x⁴ + y⁴= 110 – xy{(x² + y²− 2xy + 2xy)}

⇒ x⁴ + y⁴= 110 – xy{(x + y)²− 2xy}

xy = 103/15 and x + y = 5

⇒ x⁴ + y⁴= 110 – 103/15{(5)²− 2 × 103/15}

⇒ x⁴ + y⁴= 110 – 6.87{(25 – 13.73}

⇒ x4 + y4 = 110 – 6.87 {(11.27)}

⇒ x⁴ + y⁴= 110 – 77.42

⇒ x4 + y4 = 32.58

∴ Value of x⁴ + y⁴is 33.

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x3 + y3 = (x + y)(x2 + y2 – xy)

⇒ 22 = 5(x2 + y2 – xy)

⇒ 22 = 5[(x + y)2 − 3xy)]

⇒ 22 = 5[(5)2 − 3xy)]

⇒ xy = 103/15

(x³ + y³) (x + y) = x⁴ + y⁴ + xy(x² + y²)

(x³ + y³) (x + y)= (x⁴ + y⁴) + {xy[(x + y)² – 2xy)]

⇒ 22 × 5 = x⁴+ y⁴ + 103/15[25 - 206/15]

⇒ x⁴ + y⁴ = 32.63 ≈ 33